Question

The roots of the quadratic equation $x^2 - 6x + k = 0$ are real and distinct. How many integer values of $k$ are possible if $k$ is positive? 
 

• A. 6
• B. 7
• C. 8
• D. 9

Hint:

For quadratics, remember: real distinct roots $\Rightarrow D>0$, equal roots $\Rightarrow D=0$, complex roots $\Rightarrow D<0$.

Answer 1

The Correct Option is C

- Step 1: Condition for real and distinct roots - 

Discriminant $D > 0$. 

Here: \[ D = (-6)^2 - 4(1)(k) = 36 - 4k \] 

- Step 2: Applying the condition - \[ 36 - 4k > 0 \implies 36 > 4k \implies k < 9 \] 

- Step 3: Considering $k$ positive integer - Possible $k$: $1, 2, 3, 4, 5, 6, 7, 8$ - 8 values. 

- Step 4: Conclusion - The answer is 8, matching option (3).